First slide

Heat ;work and internal energy

Question

Two liquids at temperature 60° C and 20° C respectively have masses in the ratio 3:4 and the specific heats in the ratio 4:5, if they are mixed the resultant temperature is:

Moderate

A

70° C
B

50° C
C

40° C
D

35° C

Solution

From Principle of method of mixtures,
heat lost by hot liquid = heat gained by cold liquid.
given masses in the ratio=3:4
specific heats in the ratio=4:5
$\begin{array}{rcl} l e t h o t l i q u i d m a s s & = & 3 m, s p e c i f i c h e a t = 4 c \\ n o w c o l d l i q u i d m a s s & = & 4 m, s p e c i f i c h e a t = 5 c \\ h e a t l o s t b y h o t l i q u i d & = & m C ∆ T = 3 m (4 c) (60 - θ) \\ h e a t g a i n b y c o l d l i q u i d & = & m C ∆ T = 4 m (5 c) (θ - 20) \\ h e a t l o s t b y h o t l i q u i d & = & h e a t g a i n b y c o l d l i q u i d \\ 3 m (4 c) (60 - θ) & = & 4 m (5 c) (θ - 20) \\ \Rightarrow & θ = 35^{0} C \end{array}$

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